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Review - Part 1

Review - Part 1. Classify each item as being an allele, a genotype, or a phenotype. 1. Six fingers 2. W 3. Paired flagella 4. Kk 5. q. Review - Part 1. What are the alleles in each of these genotypes? 6. KK 7. jj 8. Qq 9. Aa. Review - Part 1.

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Review - Part 1

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  1. Review - Part 1 • Classify each item as being an allele, a genotype, or a phenotype. • 1. Six fingers • 2. W • 3. Paired flagella • 4. Kk • 5. q

  2. Review - Part 1 • What are the alleles in each of these genotypes? • 6. KK • 7. jj • 8. Qq • 9. Aa

  3. Review - Part 1 • Rewrite each of these to be a correct genotype. • 10. dD • 11. E • 12. TGtg

  4. Review - Part 1 • If R is the allele for spots and r is the allele for no spots, what will be these individuals’ phenotypes? • 13. RR • 14. Rr • 15. rr

  5. Review - Part 1 • Suppose that Y is the allele for a long tail, and y is the allele for a short tail. • 16. If zygote’s father has a short tail, what could his genotype be? • 17. If the zygote’s mother has a long tail, what could her genotype be?

  6. Review - Part 1 • What gametes will each of these individuals make? • 18. HH • 19. ii • 20. Ee

  7. Section 11-2Probability & Punnett Squares

  8. General Terms: • Zygote - A fertilized egg. New offspring. • Homozygous – genotype with two identical alleles • ex: TT – homozygous dominant or tt – homozygous recessive • Heterozygous – genotype with two different alleles • ex: Tt

  9. Practice • Identify each of these genotypes as being homozygous or heterozygous. • GG ____________ • Yy ___________ • kk ____________ • Ss ____________ • Vv ____________

  10. Practice • Identify each of these genotypes as being homozygous dominant, homozygous recessive, or heterozygous. • ee ____________ • QQ ___________ • Ll ____________ • pp ____________ • CC ____________

  11. Practice • Suppose that the I allele codes for orange fins and the i allele codes for yellow fins. • The heterozygous genotype: __ • The homozygous dominant genotype: __ • The homozygous recessive genotype: __ • A fish with yellow fins must have a _____________ genotype. • A fish with orange fins could be either _____________ or ___________________.

  12. Genetics and Probability • Figuring out offspring is a matter of chance. • Because not every gamete a parent makes is the same, which alleles are passed to the baby are random. • An example, using coins:

  13. Punnett Squares • Using the letter H to stand for heads… • If you flip a coin that’s heads (H) on both sides, what are the chances that it will come up heads (H)?

  14. Punnett Squares • If it’s a normal coin, heads (H) on one side and tails (h) on the other… • What are the odds that it will come up heads (H) on a flip?

  15. Punnett Squares • If you flip TWO normal coins, what are the odds that you will get heads (H) on both flips?

  16. Punnett Squares • The first flip will be either heads (H) or tails (h):

  17. Punnett Squares • The second flip will also be either heads (H) or tails (h):

  18. Punnett Squares • These are the possible combinations that he could have produced:

  19. Punnett Squares • These are the possible combinations that he could have produced:

  20. Punnett Squares • These are the possible combinations that he could have produced:

  21. Punnett Squares • These are the possible combinations that he could have produced:

  22. Punnett Squares • These are the possible combinations that he could have produced: H H H H H H

  23. Punnett Squares • 1 in 4 possible outcomes would be both heads (HH). 1/4 = 25% = .25 H h H H H H h H h h h h

  24. Punnett Squares • What are the odds of getting heads on one flip, tails on the other? h H H H H H h H h h h h

  25. Punnett Squares • 2 of 4 possible outcomes = 1/2 = 50% = .5 h H H H H H h H h h h h

  26. Punnett Squares • What if you flip two different coins: • One coin has two heads • The other is normal, one heads and one tails • What are the odds of getting heads on both flips?

  27. Punnett Squares H H • This is called a Punnett Square. • Punnett Squares display possible gametes and possible offspring. H H H H H H h H h h

  28. Punnett Squares • The top and side boxes show possible gametes. The middle boxes show possible zygotes (offspring) they would create.

  29. Punnett Squares • Sample problem: What are the chances that a heterozygous brown-eyed father and a homozygous recessive blue-eyed mother would have a blue-eyed child? (Use letters B/b) • Step-By-Step Instructions: • 1. Figure out Mom and Dad’s genotypes. • In the example: Dad = ___, Mom = __

  30. Punnett Squares • 2. Figure out Mom and Dad’s gametes. • Dad’s gametes = __ and __ • Mom’s gametes = __ and __ • 3. Set up a square. • For a monohybrid cross (studying only one gene), make a normal tic-tac-toe board.

  31. Punnett Squares • 4. Write Dad’s gametes on one side, and Mom’s on the other. • It doesn’t matter whether Mom or Dad is on the side vs top, just keep both eggs together and both sperm together.

  32. Punnett Squares • 4. (continued) • Like a genotype, if there’s a dominant allele, put it first. a a A a A A a A

  33. Punnett Squares • 5. Complete zygote genotypes. • Remember to put dominant allele first, if there is one. • 6. Write out all the zygote genotypes as fractions. • I.e. 1/4, 2/4, 3/4, 4/4

  34. Punnett Squares • 7. Reduce fractions if possible, and convert fractions to percentages. • For instance, if two of the four zygotes are AA, the probability of genotype AA is 2/4 = 1/2 = 50% • 8. If applicable, rewrite offspring genotype results as phenotype results.

  35. Parents – Tt and tt Offspring: Parents – TT and Tt Offspring: Practicing with Punnett Squares

  36. Practicing Punnett Squares • Show a monohybrid cross, with all genotype and phenotype probabilities, for parents who are HH and hh.

  37. Practicing Punnett Squares • Show a monohybrid cross, with all genotype and phenotype probabilities, for parents who are Ll and Ll.

  38. Practicing Punnett Squares • If Dad’s genotype is Rr and Mom is homozygous recessive, what are the odds of having homozygous dominant offspring?

  39. Practicing Punnett Squares • If both parents are heterozygous, and they have ten offspring, approximately how many of those offspring would you expect to be homozygous recessive?

  40. Practicing Punnett Squares • Suppose black fur is dominant and white fur is recessive. Two parents, one with black fur and one with white fur, have many offspring. Half of their babies are black-furred, and half are white-furred. What were the genotypes of the parents? • Hint: when a question asks you to figure out parental genotypes, make test crosses, Punnett Squares for every possibility, then see which one gives you offspring results that fit.

  41. Pedigrees • Pedigree = A diagram that shows the phenotypes of everyone in an entire family.

  42. Pedigrees • Key 7’s daughter is _. 1’s great grandson-in-law is _. Amongst 4’s nieces and nephews, _ is older than _. Both of them are _____.

  43. Section 11.3Exploring Mendelian Genetics

  44. Mendel’s Fourth Conclusion! • 4. Mendel’s Law of Independent Assortment: Genes for different traits (can) segregate independently of each other in meiosis. • Which allele from gene 1 goes into a sperm cell does not affect which allele from gene 2 the sperm will get. • Example:

  45. Mendel’s Fourth Conclusion! • If Dad has brown eyes (Bb) and a widow’s peak (Ww), his genotype is BbWw. • Half of his sperm will get the dominant B allele from the eye color gene. • But that does NOT force those sperm to also get the dominant W allele from the other gene. Some of them will get B and W, some of them will get B and w. • He will make four different kinds of gametes:BW, Bw, bW, and bw.

  46. Dihybrid Crosses • Dihybrid Cross = Punnett Square made for two genes at the same time. • Cross: RRYY X rryy [homozygous dominant round yellow peas (RRYY) and homozygous recessive wrinkled green peas (rryy)] • Result: F1 offspring were all RrYy [heterozygous round yellow peas] • Mendel let the F1 plants self pollinate: RrYy X RrYy

  47. Dihybrid Crosses 9:3:3:1 9/16 = 56% yellow round 3/16 = 19% yellow wrinkled 3/16 = 19% green round 1/16 = 6% green wrinkled In a dihybrid cross, all three laws can be observed. Law of Dominance = Rr zygotes have round seeds. Law of Segregation = Each gamete has only one allele from each gene. Law of Independent Assortment = If a gamete has a dominant allele from one gene, it doesn’t have to also have a dominant allele from the other gene.

  48. Patterns of Inheritance • So far, we’ve only studied simple dominance, where a gene has two alleles, one dominant and one recessive. • Most genes aren’t like that, and have more complicated patterns of inheritance. • You must be able to recognize five more advanced patterns on sight, and be able to make monohybrid crosses for three of them.

  49. Patterns of Inheritance • Polygenic Traits – The phenotype is determined by more than one gene at once. • Example: Eye color, in reality, doesn’t just come from one B/b gene. There’s a gene for blue vs green tint, a gene for dark vs light, a gene for tan vs gold tint, a gene for color density, a gene for outlying vs inlying rings… and your exact eye color depends on all of them.

  50. Patterns of Inheritance • Multiallelic Trait – A single gene has more than two possible alleles. • Example: In cats, coat color is partly determined by the browning gene, which has THREE alleles. B is the allele for black, b is the allele for chocolate brown, and bl is the allele for cinnamon brown. • Possible genotypes for this gene are: BB, Bb, Bbl, bb, blbl, bbl. • If you cross a Bb cat with a blbl cat, what are the offspring genotype probabilities?

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